bs and ec

British Standard vs Eurocode 3 – Steel Building Design

A comparison between BS 5950 and Eurocode 3. BS 5950 Eurocode 3 Terminology Force Capacity Mc Design strength py Dead load Live load Wind load

Action Resistance Mc,Rd Yield strength fy Permanent load Variable load Another variable load

Symbol Elastic Modulus Z Wel Plastic Modulus S Wpl Radius of Gyration r i Torsion constant J It Warping constant H Iw Changes in load factor 1.4Gk + 1.6Qk 1.35Gk + 1.5 Qk

Classification: 5.0)275(yp

=ε Classification: 5.0)235(yf

=ε

Moment Resistance Class 1 and 2: Mc = pyS

Class 1 and 2: Mc,Rd = 1M

plyWfγ

Tu Trung Nguyen – MSc. Structural Engineering


Class 3 semi-compact: Mc = pyZ or Mc = pySeff Class 3: Mc,Rd =

1

min,

M

elyWfγ

Class 4: slender Mc = pyZeff Class 4: Mc,Rd =

1

min,

M

effyWfγ

Low shear Fv < 60% Pv

Low Shear VEd < 50% Vpl,Rd

Shear Resistance Pv = 0.6pyA

VEd = 1

3M

yv

fA

γ

Shear Area Av = tD

Shear Area Av = A – 2btf – (tw + 2r)tf 1.04tD = htw ≈

Shear bucking d/t > 70ε

Shear bucking hw/tw > 72ε

Deflection - Serviceability LS Imposed load only Permanent action, 1δ

Variable action, 2δ Pre-camber, 0δ

Span/360 – brittle Span/200 - general

maxδ < L/250

2δ < L/350 – brittle

3δ < L/300 – general

Compression Members Pc = Agpc from table 23, 24 BS 5950 pc is a function of λ Nb,Rd =

1M

ya Afγ

χβ

χ : reduction factor depends on λ , non-dimensional slenderness

λ = λ−Φ+Φ 2

1 1≤

])2.0(1[5.02

λλ +−+=Φ a

λ = 1λλβ

=cr

ya

NAf

; yf

Eπλ =1

Ncr = cr

eff

LEI2

2π; Ieff = 0.5 IAho μ22 +

μ from table 6.8



LTB Mx < Mb/mLT and Mx < Mcx Mb = pb x modulus pb from LTλ from table 16

LTλ = wu βνλ (see below)

y

E

rL

=λ

ry = the radius of gyration about the minor axis Class 1 and 2: wβ = 1.0, Mb = pbSx

Class 3 semi-compact: Mc = pyZ => wβ = Zx/Sx Mc = pySeff => wβ = Sx,eff/Sx Class 4 slender cross-sections:

wβ = Zx,eff/Sx mLT from Table 18

Equal flange: 25.02 ])/(05.01[1

xλν

+=

u = 0.9 , x = D/T equal flange, I and H

1.0 for UB, UC same approach as for compression

Mb,Rd = 1

,

M

yyplaLT fWγβχ

LTλ = LTLTLT λ−Φ+Φ 2

1 1≤

])2.0(1[5.02

LTLTLTLT a λλ +−+=Φ

LTλ = 1

.

λλ

=cr

yypl

MfW

; yf

Eπλ =1

Mcr = z

t

z

wZ

EIGIL

II

LEIC 2

2

2

2

1 ππ

+

C1 results from bending diagram below aLT = 0.34 for rolled UC section aLT = 0.49 for rolled UB section Approx:

• Mcr = 22 )(

20119.0

hiLt

iL

Ah

z

f

z

+

⎟⎟⎠

⎞⎜⎜⎝

⎛

Quick determine LTλ , using LTλ = wu βνλ where y

E

rL

=λ , conservative: u = 0.9, v=1, wβ =1



Eq. 5.14 EC3 cl.5.3.3(1,2,3)

NEd = MEd/h Deflection:

e0 = 500

Lam ; am = )11(5.0m

+

m: number of members to be restrained

1. Load factor yM0 = 1,00, yM1 = 1,00, yM2 = 1,25

2. Tension:

1,

≤Rdt

Ed

NN ; Nt,Rd = min (

20

9.0;M

uNet

M

y

yfA

yAf

) EC3:2005 cl.6.2.3 (1,2)



Where: fu = 1.5fy (cl.3.2.2 (1))

Anet = A – (number of bolts)x(Diameter of bolts) cl.6.2.2.2

3.Compression

1,

≤Rdc

Ed

NN where Nc,Rd =

0M

y

yAf

for class 1,2 and 3 cl.6.2.4

4. Bending moment

1,

≤Rdc

Ed

MM cl.6.2.5

Class 1, 2 Class 3 Class 4

Mc,Rd =0

,M

yplRdpl y

fWM = Mc,Rd =

0

min,,

M

yelRdel y

fWM = Mc,Rd =

0

,

M

yeffel

yfW

Shear cl.6.2.6

1,

≤Rdpl

Ed

VV where Vpl,Rd =

0

3M

yv

y

fA



5.Bending and Shear Cl.6.2.8

N = yf)1( ρ− where 2

,

)12( −=Rdpl

Ed

VVρ

My,V,Rd = RdcM

yw

wypl

My

ft

AW

,0

2

, ]4

[≤

−ρ

Where Aw =hwtw

6.Bending and axial force Class 1, and 2 cl.6.2.9.1 MEd < MN,Rd

+ For only rectangular section

MN,Rd = Mpl,Rd [1- 2

,

)(Rdpl

Ed

NN ]

+ For doubly symmetrical I and H section or other flange section to resistance about y-y NEd < 0.25Npl,Rd



NEd < 0

5.0

M

yww

yfth



7.Guide for design curved beam to Eurocode 3 Determine out of plane bending stress:

Direct stress AN

WM

el

+=1σ Out-of-plane bending stress: RT

B21

23σσ =

R: radius of the section, T: Thickness of flange. B is outstand of the flange

B = )2(21 rtb −−

Reduce design strength

fyd = 2

3)2

(3 25.0

2222 στσ+⎥⎦

⎤⎢⎣⎡ −−yf

- Check axial force (Please see Section 3) - Check Bending capacity (please see Section 4) - Check section capacity under axial load and bending moment at Section 6 (more obviously) Briefly with class 1 and 2: I and H section MEd < MN,z,Rd

n = Rdpl

Ed

NN

,

a = (A-2btf)/A < 0.5

+ n < a MN,Rd = Mpl,z,Rd

+ n > a MN,z,Rd = Mpl,z,Rd [1+ 2)1

(aan

−− ] = Mpl,z,Rd )

5.011(

an

−−

- Check member buckling (cl.6.3.3)



Reference: - Eurocode 3:1993-1-1:2005: General rules for steel building design - Bristish Standards 5950-1:2000: Code of practice for design rolled and welded sections - Charles King and David Brown (2001) Design for curved beam, SCI


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